in PROBABILITY
A CONNECTION BETWEEN THE STOCHASTIC HEAT EQUATION
AND FRACTIONAL BROWNIAN MOTION, AND A SIMPLE PROOF
OF A RESULT OF TALAGRAND
CARL MUELLER1 Dept. of Mathematics University of Rochester Rochester, NY 14627
email: [email protected]
ZHIXIN WU
Dept. of Mathematics DePauw University Greencastle, IN 46135
email: [email protected]
SubmittedSeptember 10, 2008, accepted in final formJanuary 8, 2009 AMS 2000 Subject classification: Primary, 60H15; Secondary, 35R60, 35K05. Keywords: heat equation, white noise, stochastic partial differential equations.
Abstract
We give a new representation of fractional Brownian motion with Hurst parameterH ≤ 1 2 using stochastic partial differential equations. This representation allows us to use the Markov property and time reversal, tools which are not usually available for fractional Brownian motion. We then give simple proofs that fractional Brownian motion does not hit points in the critical dimension, and that it does not have double points in the critical dimension. These facts were already known, but our proofs are quite simple and use some ideas of Lévy.
1
Introduction
Our main result is a new representation for fractional Brownian motion using stochastic partial differential equations, described in this section. As an application, in Section 2, Theorems 1 and 2, we state some known results about when fractional Brownian motion hits points and has double points. Our representation allows us to give simple proofs of these results.
In recent years there has been an upsurge of interest in fractional Brownian motion, see Nualart, Chapter 5 [Nua06]. The most common model for noise in physical systems is white noise ˙Bt, the derivative of Brownian motion. The central limit theorem gives some justification for using a Gaussian process such asBt. Furthermore, ˙Bs, ˙Bt are independent ifs6=t. In many situations, however, there are correlations between noise at different times. A natural correlated Gaussian
1RESEARCH SUPPORTED BY THE NSF AND NSA
model to consider is fractional Brownian motionXt =XtH :t≥0 taking values inR
n, with Hurst parameterH∈(0, 1]. The processXt is uniquely specified by the following axioms.
1. X0=0 with probability 1.
2. Xt :t≥0 is a Gaussian process with stationary increments. That is, for t,h>0 the proba-bility distribution of the incrementXt+h−Xt is independent oft.
3. Forc>0 we haveXc t=D cHXt, where=D denotes equality in distribution. 4. X1has the standard normal distribution inRn.
Note that Brownian motion is a fractional Brownian motion with Hurst parameterH=1/2. Next, we describe a seemingly unrelated process, the solution of the heat equation with additive Gaussian noise. Then we show that fractional Brownian motion can be recovered from this so-lution. There are several representations of fractional Brownian motion, see Nualart [Nua06], Chapter 5. One advantage of our representation is that we can use the Markov property and time reversal, tools which fail for the fractional Brownian motion alone. Using these extra tools, we give a simple proof of some hitting properties of fractional Brownian motion, and a result of Talagrand about double points. Throughout the paper we will write SPDE for “stochastic partial differential equation”.
Informally speaking, we wish to consider solutionsu(t,x):(t,x)∈R×Rnto the following equa-tion.
∂tu = ∆u+F(t,˙ x) u(−∞,x) = 0
where ˙F(t,x)is a generalized Gaussian field with the following covariance: EF˙(t,x)F˙(s,y)=δ(t−s)h(x−y) where
h(x) = ¨
δ(x) ifH=1 2 |x|−α otherwise and
H=1−αn 2 so that
α= 1−2H
n .
We will require
0<H≤1 2 so that
0≤α <1
n. Note that ifH> 1
2 thenα <0 andh(0) =0, and thenhis not a proper covariance. Our goal is to show thatXt
D
The above description is not rigorous. To be precise, ˙F is a centered Gaussian random linear functional on C∞c (Rn+1), the set of infinitely differentiable functions with compact support on (t,x)∈R×Rn, with covariance
Q(f,g):=EF(f)F(g)= Z
R
Z
Rn
f(t,x)g(t,x)d x d t (1.1)
ifH=1−n 2 , and
Q(f,g):=EF(f)F(g)= Z
R
Z
Rn
Z
Rn
f(t,x)g(t,y)h(x−y)d y d x d t (1.2)
ifH6= 1−n
2 . Note that in either case, the integral in (1.1) or (1.2) is nonnegative definite. Thus, we can extendF(f)to all functionsf satisfying
Q(f,f)<∞.
We call this class of functionsH. Note thatHimplicitly depends onα,n. Next, fort>0 andx∈Rnlet
G(t,x):= (
(4πt)−n/2exp−|x|2
4t
ift>0
0 ift≤0
be the heat kernel onRn.
We would be tempted to defineu(t,x)by
u(t,x) = Zt
−∞ Z
Rn
G(t−s,x−y)F(d y ds)
but the integral will not converge. However,u(t,x)−u(0, 0)looks more promising. WhenH=1/4 the stationary pinned string was defined in[MT02]as
U(t,x):= Z t
−∞ Z
Rn
G(t−s,x−y)−G(−s,−y)F(d y ds). (1.3)
This definition also works for other values of H, provided g ∈H, where g(s,y) = gt,x(s,y):= G(t−s,x−y)−G(−s,−y).
Lemma 1. Let g be as in the previous paragraph. For all t>0we have g(s,y)1(s≤t)∈H.
We will prove Lemma 1 in the Appendix.
From the covariance of ˙Fone can easily deduce the following scaling property. We leave the proof to the reader.
Lemma 2. The noiseF obeys the following scaling relation,˙ ˙
Turning to the SPDE, definev(t,x)byav(t,x) =U(c t,c1/2x). The reader can verify the following calculation using (1.3).
a∂tv = c∂tU
= c∆U+F˙(c t,c1/2x) D
= a∆v+c1−(1+αn)/2F˙(t,x) D
= a∆v+c(1/2)−(αn/2)F˙(t,x)
where the equality in distribution holds for the entire random field indexed byt,x. Thus, we can cancel out the constantsc,aprovided
a=c1−αn2
and thenvsatisfies the same equation asu. Thus,
aU(t,x)=D U(c t,c1/2x). Setting x=0 gives us the scaling relation forU(t, 0). Thus we find Lemma 3. U(t, 0)obeys the following scaling relation. For c>0we have
U(c t, 0)=D c1−αn2 U(t, 0)
where the equality in distribution holds for the entire process indexed by t.
Remark 1. Let Vt,x(s,y) =U(t+s,x+ y)−U(t,x). It follows immediately from (1.3) that the random fields Vt,x(s,y)and U(s,y)are equal in distribution.
Let
Xt=KαU(t, 0)
where
Kα=
(2−α)Γ( n 2) 2−3α2+1Γ(n−α
2 )
1/2
ifα6=1
and
Kα=2−1/2(4π)d/4ifα=1.
We claim that
Proposition 1. Assume that0≤α < 1
n, and let Xt =KαU(t, 0). Then Xt, as defined above, is a fractional Brownian motion with Hurst parameter
H=1−αn 2 .
Proof. We only need to verify the four axioms for fractional Brownian motion. It follows from (1.3) thatX0=0, so axiom 1 is satisfied. Axiom 2 follows from Remark 1. Axiom 3 follows from the scaling properties of fractional Brownian motion and Lemma 3. Finally, Axiom 4 follows from (1.3) and the integral of the covarianceh, which we verify in the Appendix.
2
Critical dimension for hitting points, and for double points
The rest of the paper is devoted to the following questions. 1. For which values ofd,HdoesXt hit points?
2. For which values ofd,HdoesXt have double points?
Recall that we sayXt hits points if for eachz∈Rn, there is a positive probability thatX
t =z for some t>0. We say that Xt has double points if there is a positive probability thatXs =Xt for some positive timest6=s. Here are our main results.
Theorem 1. Assume0<H≤1
2, and that 1
H is an integer. For the critical dimension n= 1
H, fractional Brownian motion does not hit points.
Theorem 2. Assume0<H≤12, and that 2
H is an integer. For the critical dimension n= 2
H, fractional Brownian motion does not have double points.
In fact, Talagrand answered the question of double points in[Tal98], Theorem 1.1, and the as-sertion about hitting points was already known. Techniques from Gaussian processes, such as Theorem 22.1 of[GH80], can usually answer such questions except in the critical case, which is much more delicate. The critical case is the set of parametersn,H which lie on the boundary of the parameter set where the property occurs. For example, n=2,H = 1
2 falls in the the critical case for fractional Brownian motion to hit points. But forH= 1
2we just have standard Brownian motion, which does not hit points inR2. This illustrates the usual situation, that hitting does not occur, or double points do not occur, in the critical case.
It is not hard to guess the critical parameter set for fractional Brownian motion hitting points or having double points. Heuristically, the range of a process with scaling Xαt
D = αHX
t should have Hausdorff dimension 1
H, ifXt takes values in a space of dimension at least 1
H. For example, Brownian motion satisfiesBαt =D α1/2B
t, and Brownian motion has range of Hausdorff dimension 2, at least if the Brownian motion takes values in Rn withn ≥ 2. The critical parameter ofH for a process to hit points should be when the dimension of the range equals the dimension of the space. Thus, the critical case for fractional Brownian motion taking values inRn should be when the Hurst parameter isH =1/n. For double points, we consider the 2-parameter process V(s,t) =Xt−Xs. This process hits zero at double points ofXt, except whent=s. The Hausdorff dimension of the range ofV should ben= 2
H, and so the critical Hurst parameter for double points ofXtshould beH= 2n1.
First note that the supercritical case can be reduced to the critical case. That is, ifXt= (X
(1)
t ), . . . ,X
(n+m)
t ) = 0, then it is also true that the projection(X(t1)), . . . ,X(tn)) =0. Furthermore, the subcritical case is easier than the critical case, and it can be analyzed using Theorem 22.1 of Geman and Horowitz
[GH80]. Therefore, we concentrate on the critical caseH=1/n.
3
Summary of Lévy’s argument
Here is a brief summary of Lévy’s argument that 2-dimensional Brownian motion does not hit points. Letm(d x)denote Lebesgue measure onRnand letB
t denote Brownian motion onRn. For this section, letn=2. Furthermore, letB[a,b]:={Bt :a≤t≤b}. It suffices to show that
since then we would have
0 = E
Recall thatYt,Zt : 0≤t≤1 are independent standard 2-dimensional Brownian motions. This is a standard property of Brownian motion, which can be verfied by examining the covariances of Yt,Zt: 0≤t≤1. ThenY[0, 1],Z[0, 1]are independent random sets. Furthermore, by Brownian On the other hand, set theory gives us
By the independence ofY[0, 1]andZ[0, 1]and the Cauchy-Schwarz inequality, we have
0 = Z
R2
E1Y[0,1](z)1Z[0,1](z)
dz (3.3)
= Z
R2
E1Y[0,1](z)E1Z[0,1](z)dz
= Z
R2
E1Y[0,1](z)
2 dz
≥ Z
R2
E1Y[0,1](z)dz 2
=
Em(Y[0, 1]) 2
.
ThereforeE[m(Y[0, 1])] =0 and (3.1) follows from the definition ofY.
4
Proof of Theorems 1 and 2
Now we use Lévy’s argument to prove our main theorems.
4.1
Hitting points, Theorem 1
The argument exactly follows that in Section 3, except thatR2is replaced byRn. Also, by axiom (3),
Xc t D =cHXt
forc>0. However, sinceXt takes values inRnandH=1/n, we still have E
h
m(X[0, 2]) i
=2E h
m(X[0, 1]) i
=2E h
m(X[1, 2]) i
.
Recall thatm(·)denotes Lebesgue measure inRn. As before, let Yt = X1+t−X1 Zt = X1−t−X1.
It is no longer true that Y[0, 1],Z[0, 1]are independent. Now we use the fact that Xt is equal in distribution tou(t, 0), whereu(t,x)is the stationary pinned string. Changing the probability space if necessary, let us writeXt=KαU(t, 0), and letHtdenote theσ-field generated byU(t,x):
x∈Rn. Then we have the following lemma.
Lemma 4. Let us use the above notation. Then Y[0, 1],Z[0, 1]are conditionally independent and identically distributed givenH1.
Proof of Lemma 4. The lemma is proved in[MT02], Corollary 1, forH =1/4, and the proof for other values ofH uses similar ideas.
are equal in distribution. and thereforeYt,Zt are identically distributed processes.
Next we discuss the conditional independence ofY[0, 1],Z[0, 1]givenH1. First we claim that the stationary pinned stringU(t,x)enjoys the Markov property with respect tot. This is a general fact about stochastic evolution equations, and we refer the reader to [Wal86], Chapter 3. It follows thatZ[0, 1]is conditionally independent ofY[0, 1]givenH1.
This proves Lemma 4.
From here we duplicate the argument in Section 3, replacing expectation by conditional expecta-tion givenH1. Briefly, it suffices to show that
E
But, following the same argument as before, we conclude that with probability one, E
We leave it to the reader to verify that (3.2) and (3.3) still hold, provided expectation is replaced by conditional expectation givenH1. This verifies (4.1), and finishes the proof of Theorem 1.
4.2
Double points, Theorem 2
To show that Xt does not have double points in the critical case, we use the same argument, but applied to the two-parameter process
V(s,t):=X(t)−X(s).
We need to show thatV(s,t)has no zeros except ifs=t. To simplify the argument, we will show thatV(s,t)has no zeros for(s,t)∈ R, where
The same argument would apply to any other rectangle whose intersection with the diagonal has measure 0. Let us subdivideRinto 4 subrectanglesRi:i=1, . . . , 4 each of which is a translation of[0, 1]2. Again we argue as in Section 3. By scaling, we see that for eachi=1, . . . , 4
E[m(V(R))] =4Em(V(Ri)).
Next, letH1be theσ-field generated by{u(1,x):x ∈Rn}, and suppose we have labeled theRi such thatR1= [0, 1]×[4, 5]andR2= [1, 2]×[4, 5]. Thus, as before, for each pairi 6= j ∈ {1, . . . , 4}we have
E h
m(V(Ri)∩V(Rj)) ¯ ¯
¯H1i=0.
Now in[MT02], Corollary 1, it was shown that forH=1/4,V(R1),V(R2)are conditionally i.i.d. givenH1. For other values ofH, the argument is very similar to the proof of Lemma 4, and we leave the details to the reader.
Therefore, as in Section 3, we conclude that with probability one,
E h
m(V(Ri)) ¯ ¯
¯H1i=0.
Also as in Section 3, this finishes the proof of Theorem 2.
A
Appendix
Proof of Lemma 1. In the case ofH=1
2, whereh(x) =δ(x)is not difficult to show that Zt
0 Z
Rn
G(r,x−y)2d y d r=2(4π)−d/2t1/2
and we leave it to the reader. Moving on to the case ofH6= 1−n
2 , we calculate the following integral. Z
Rn
Z
Rn
G(r,x−z)G(r,x−z′)|z−z′|−αdzdz′
We make use the Fourier transform, with the following definition.
ˆ f(ξ) =
Z
Rn
f(x)ei x·ξd x
forx,ξ∈Rn.
Iff(x−z) =G(r,x−z),g(x−z′) =G(r,x−z′),h(z−z′) =|z−z′|−α,H(z) =g∗h(x−z) =J(x−z),
then
ˆ H(z) =
Z
Rn
Therefore, It is known that
ˆ
by (A.2) and (A.3), (A.1) becomes Z
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